Calculate the binding energy of `""_(3)^(6)Li` assuming the mass of `""_(3)^(6)Li` atom as 6.01512 amu.

Explanation

Mass defect is
`Deltam = [Zm_(p) + (A-Z) m_(n) -M]`…………(1)
`m_(p)` = mass of the proton = 1.007277 amu
`m_n` = mass of the neutron = 1.008655 amu
Mass of the bound =6.01512 amu
[We know as per the symbol, lithium has three protons and three neutrons, since Z = 3, A = 6, and number of neutrons = A- Z = 6- 3 = 3]
Substituting this in Eq. (1), we have-
`Deltam =[3 xx 1.007277 + 3(1.008655)]-6.01512`
`Deltam =[3.02183+ 3.02596]-6.01512`
`=6.047795 - 6.01512 = 0.032675` amu.
But the energy equivalent of 1 amu = 931 MeV Therefore, the binding energy which is equal to the energy equivalent of mass defect in amu is
`E_(b) = Delta xx 931 MeV`
`=0.032675 xx 931 = 30.42 MeV`